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This section includes InterviewSolutions, each offering curated multiple-choice questions to sharpen your knowledge and support exam preparation. Choose a topic below to get started.
| 41901. |
The photoelectric cut-off voltage in a certain experiment is 1.5 v. What is the maximum kinetic energy of photoelectrons emitted? |
| Answer» SOLUTION :`1.5eV=2.4xx10^(-19)J` | |
| 41902. |
If two capacitors C_(1) and C_(2) are connected in a parallel combination then the equivalent capacitanceis 10 mu F . If both the capacitors are connected across a l V battery, then energy stored in C_(2) is 4 times of that in C_(1). The equivalent capacitance if they are connected in series is |
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Answer» `16 MU F `  For parallel connection and from QUESTION `4 ((1)/(2) C_(1)^(2))= (1)/(2) C_(2)V^(2)` `:. 4 C_(1)= C_(2)` By using EQUATION (2) in equation (1), `C_(1) + 4 C_(1) = 10 mu F ` ` :. 5 C_(1) = 10 mu F` `:. C_(1) = 2 mu F ` From equation (2) `C_(2) = 4 C_(1)` `= 4xx2` `= 8 mu F ` Now, equivalent capacitance of series connection of `2mu`F and `8mu`F `C = (C_(1)C_(2))/(C_(1)+C_(2))=(2xx8)/(2+8)= (16)/(10)` `= 1.6 mu F ` |
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| 41903. |
Why choke coil cannot be used in dc ? |
| Answer» Solution :Since the FREQUENCY of DC is zero , the inductive reactance offered by the CHOKE `X_L=Lomega=0`. So it cannot be used in dc. | |
| 41904. |
The pole strength of a bar magnet is 100 units. It is cut into two pieces along a line parallel to its equator in such a way that the length of one piece is half the length of the other. What will be the ratio of the pole strengths of the two pieces? |
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Answer» `1 : 1` `THEREFORE` The ratio of the pole strengths of the TWO pieces will be 1 : 1. However, the magnetic moments of the two pieces will be different. |
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| 41905. |
What is the resolving power of a telescope, for light of wavelength 5000 A, if the aperture of the objective of the telescope is 1.22 m ? |
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Answer» `1.50 XX 10^6` |
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| 41906. |
The speed of the earth's rotation about it's axis is omega. Its angular speed is increased to make the effective acceleration due to gravity equal to zero at the equator. Then what will be the final omega. |
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Answer» `sqrt((g)/(R))` |
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| 41907. |
Nature of electric force between two protons is: |
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Answer» Attractive |
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| 41908. |
Two bullets are fired simultaneously, horizontally and with different speeds from the same place. Which bullet will hit the ground first? |
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Answer» the FASTER ONE |
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| 41909. |
The frequencies of X-ray, gamma-rays and ultraviolet rays are respectively a, b, and c. Then : |
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Answer» `a LT B, b lt C` |
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| 41910. |
Therefore are 6xx10^(19) electrons per unit cubic metre of pure semiconductor. What will be the number of holes for this semiconductor of dimension 1 cm xx 1 cm xx 2 cm? |
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Answer» `6xx10^(19)` VOLUME of crystal `v=l xx b xx h` `=1xx1xx2xx10^(-6)m^(3)` `=2xx10^(-6)m^(3)` In PURE semiconductor `n_( e)=n_(h)=6xx10^(19)m^(-3)` `THEREFORE` Total number of holes in crystal, `N=n_( e)V=6xx10^(19)xx2xx10^(-6)=12xx10^(13)` |
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| 41911. |
Obtain the expression of electric field at any point by continuous distribution of charge on a volume. |
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Answer» Solution :Suppose a continuous charge DISTRIBUTION in space has a charge density `rho`. Choose any convenient origin O and let the position vector of any point in the charge distribution be `vecr`.  Divide the charge distribution into small volume elements of size `DeltaV`. The charge in a volume element `DeltaV` is `rhoDeltaV`. Now, CONSIDER any general point P (inside or outside the distribution with position vector `vecR`. Electric field due to the charge `rhoDeltaV`is given by Coulomb.s LAW. `vecE = sum(krhoDeltaV)/(r.)^(2).hatr` where r. is the distance between the charge element and P and f. is a unit vector in the direction from the charge element to P. By the superposition principle, the total electric field, `vecE = int_(V) (krho.DeltaV)/(r.)^(2)hatr = k -= (rho DeltaV)/(r^(2)).hatr` OR `vecE = kint(rho.DeltaV)/(r^(-2).hatr` In short, using Coulomb.s law and the superposition principle, electric field can be DETERMINED for any charge distribution, discrete or continuous or part discrete and part continuous. |
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| 41912. |
A spectral line caused by the transition .^(3)D_(1) rarr ^(3)P_(0) experiences the Zeeman splitting in a weak magnetic field. When observed at right angles to the magnetic field direction, the interval between the neighbouring components of the split line is Delta omega= 1.32.10^(10) s^(-1). Find the magnetic field induction B at the point where the source is located. |
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Answer» Solution :The `.^(3)P_(0)` term does not split. The `.^(3)D_(1)` term splits into 3 lines corresponding to the shift. `DELTAE= -gmu_(B)BM_(Ƶ)` with `M_(Ƶ)=+- 1,0`. The interval between neighbouring components is then GIVEN by `ħ DELTA omega= g mu_(B)B` HENCE `B=(ħDelta omega)/(gmu_(B))` Now for the `.^(3)D_(1)` term `g=1+(1xx2+1xx2-2xx)/(2xx1xx2)=1+(4-6)/(4)=(1)/(2)` Substitution gives `B= 3.00KG= 0.3T`. |
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| 41913. |
The graph gives the magnitude B (t) of a uniform magnetic field that exists throughout a conducting loop, perpendicular to the plane of the loop. Rank the five regions of the graph according to the magnitude of the emf induced in the loop, greatest first |
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Answer» `b gt (d=e) LT (a=C)` |
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| 41914. |
The wave number of two waves are 2.5 xx 10^6 //metre and 2 xx 10^6 //metre. What is the difference between their wavelength ? |
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Answer» 5000 a |
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| 41915. |
The magnetic flux linked with a coil varies as phi=3t^(2)+4t+9. Find the magnitude of the emf induced at t = 2S. |
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Answer» SOLUTION :Induced emf =`-(dphi)/(dt)` `=-d/(dt)(3t^2 + 4t+9)` =-(6t+4) VOLT Induced emf at t =2 , e=-(6 x 2 +4) =-16 V |
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| 41916. |
A : Strikes uses glasses. R : Light reflected by snow is partially polarised. |
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Answer» Both A and R are true and R is the correct explanation of A |
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| 41917. |
One end of a glass rod 8cm in diameter has a hemishperical surface of 4cm radius. Determine the position of the image of an object placed on the axis at following distance from the hemishperical end (a) at infinity, (b) 16cm, (c ) 4cm, refractive index of the glass is 1.5. |
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Answer» |
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| 41918. |
A current i flows through an infinitely long wire having infinite bands as shown. The rad ius of the first curved section is a and the radii of the successive curved portions each increases by a factor eta . Find magnetic field at O. |
| Answer» Solution :`B = ((mu_0 itheta)/(2 PIA))/(1 + 1/eta)=(mu_0 itheta eta)/(4 PI a(eta+ 1))` | |
| 41919. |
When a train is at a distance of 2 km , its engine sounds a whistle . A man near the railway track hears the whistle directly and by placing his ear against the track of thetrain. If the two sounds are heard at an internal of 5.2 s, the speed of the sound in iron ( material of the rail track ) is : (Given that velocity of sound in air is 330 m s ^(_1) ) |
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Answer» `2,325.6 MS^(-1)` |
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| 41920. |
Light of wavelength 620 nm passes through a double slit, yielding a diffraction pattern whose graph of intensity l versus angular position theta is shown in Figure. Calculate (a) the slit width and (b) the slit separation. (c) Verify the displayed intensities of thc m = 1 and m = 2 interference fringes. |
| Answer» Solution :(a) 7.1 `MU m`, (B) 28 `mu m`, (c) for m=1, `I= 5.7 mW // CM^(2)`, for m = 2, `I = 2.9 mW // cm^(2)` | |
| 41921. |
In Rutherford experiment , the number of the alpha particles scattered through and angle of 60^@ by a silver foil is 200 per minute. When the silver foil is replaced by a copper foil of the same thickness , the number ofalpha- particles scatted through an angle of 60^@ per minute is : |
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Answer» `200 XX Z_(Cu)/(Z_(Ag))` |
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| 41922. |
In the circuit shown in figure q varies with time t as q=(t^(2)=16). Here q is in coulomb and t in second. Find V_(ab) "at" t=5s |
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Answer» 50 V |
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| 41923. |
In the circuit shown in figure q varies with time t as q=(t^(2)=16). Here q is in coulomb and t in second. Find V_(ab)=(V_(a)-V_(b)) at t=3s |
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Answer» `-24.5 V` |
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| 41924. |
WHO WAS NANUKAKA? |
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Answer» MOTHER'S BROTHER |
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| 41925. |
In a hydrogen atom, an electron of mass m and charge e revolves in an orbit of radius r making n revolutions per second. If the mass of hydrogen nucleus is M, the magnetic moment associated with the orbital motion of electron is |
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Answer» `(pi N E R^(2) m)/(M+m)` |
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| 41926. |
The donor impurity to be added for doping germanium crystal, will be of valency |
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Answer» 2 |
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| 41927. |
Assertion: the ferromagnetic substance do not obey Curie's Law. Reason : At Curie point a ferromagnetic substance start behaving as a paramgnetic substance. |
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Answer» If both assertion and reason are true and reason is the correct EXPLANATION of assertion. Explanation : the susceptibility of ferromagnetic substance DECREASES with the rise of temperature in a complicated MANNER. After Curie POINT the susceptibility ferromagnetic substance varies inversely with its ABSOLUTE temperature. |
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| 41928. |
Four capacitors marked with capacitances and breakdown voltages are connected as shown in the figure. The maximum emf of the source, so that no capacitor breaks down is |
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Answer» 10.5kV  Resultant capacitance of SERIES combination 1, `=(1)/(5) + (1)/(4) = (9)/(20)` `C_(eq_(1)) = (20)/(9) = 2.25muF` Resultant capacitance of series combination 2. `=(1)/(3) + (1)/(2) = (5)/(6)` `C_(eq_(2)) = (6)/(5) = 1.2 MU F` so, charge on upper BRANCH is `= (20)/(9)V` and charge on lower branch is `(6)/(5)V`. 
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| 41929. |
The inverse square law in electrostatics is |F|=(e^(2))/((4pi in_(0)r^(2)) for the force between an electron and a proton. The (1/r) dependence of |F| can be understood in quantum theory as being due to the fact that the 'particle' of light (photon) is massless. If photon had a mass m_(p), force would be modified to|F|=(e^(2))/((4pi in_(0))) [1/(r^(2))+lambda/r]. Exp (-lambdar) where lambda =m_(p)c//h and h=h/(2pi) Estimate the change in the ground state energy of a H-atom if m_(p) were 10^(-6) times the mass of an electron. |
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Answer» Solution :Here, we suppose MASS of photon, `m_(p)=10^(-6)` electronic mass i.e., `m_(p)=10^(-6)(0.5)MeV=5xx10^(-7)xx1.6xx10^(-13)J=0.8xx10^(-19)J`. Now, `1/(lambda)=h/(m_(p)c)=(hc)/(m_(p)c^(2))=(10^(-34)xx3xx10^(8))/(0.8xx10^(-19))=4xx10^(-7)m` which is much larger than Bohr radius. As `|vecF|=(e^(2))/(4pi in_(0))[1/(r^(2))+lambda/r] exp. (-LAMBDAR).....(i) where lambda^(-1)=h/(m_(p)c)=4xx10^(-7)mgt gt r_(B)` `:. lambdalt lt1/(r_(B)) or lambda_(B)lt lt1` Now, `U(r)=(-e^(2))/(4pi in_(0)) (exp (-lambdar))/r......(II)` As `mvr=h. v=h/(mr). As exp (-lambdar)to1`, therefor, form (i) ` (mv^(2))/r=F=(e^(2))/(4pi in_(0))[1/(r^(2))+lambda/r]` `:. m/r (h/(mr))^(2)=(e^(2))/(4pi in_(0))[1/(r^(2))+lambda/r]` `(h^(2))/m=(e^(2))/(4pi in_(0)) r+lambdar^(2)........(iii)` If `lambda=0, r_(B), then (h^(2))/m=(e^(2))/(4pi in_(0))r_(B)`. As `lambda^(-1)gt gt r_(B)`, put `r=r_(B)+delta` form (iii), `r_(B)=(r_(B)+delta)+lambda(r_(B)+delta)^(2)=r_(B)+delta+lambda(r_(B)^(2)+delta^(2)2delta r_(B))` Neglecting `delta^(2)`, we get `0=lambda r_(B)^(2)+delta(1+2lambda r_(B)^(2))` `delta=(-lambdar_(B)^(2))/(1+2lambdar_(B))=-lambdar_(B)^(2)(1-2lambda r_(B))` `delta=-lambdar_(B)^(2)(As lambdar_(B)lt lt1)` form (ii), `V(r)=(-e^(2))/(4pi in_(0)) (exp. (-lambda delta-lambdar_(B)))/(r_(B)+delta)=(-e^(2))/(4pi in_(0)) 1/(r_(B))[(1-delta/(r_(B))) (1-lambdar_(B))]=-27.2eV` i.e., V(r) remai ns unchanged. `K.E.=1/2mv^(2)=1/2m(h/(mr))^(2)=(h^(2))/(2m(r_(B)+delta)^(2)) = (h^(2))/(2mr_(B)^(2)) (1-(2delta)/(r_(B)))=(13.6eV)[1+2lambdar_(B)]` Total energy `=(-e^(2))/(4pi in_(0)r_(B))+(h^(2))/(2mr_(B)^(2)) (1+2lambda r_(B))=-27.2+13.6(1+2lambdar_(B))eV.` Change in ground state energy of hydrogen atom `=13.6xx2lambda_(B)eV=27.2lambdar_(B)eV` |
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| 41930. |
Define magnetic flux density. |
| Answer» Solution :The magnetic flux density is defined as the NUMBER of magnetic FIELD lines crossing unit area kept normal OT the direction of line of force . Its unit is WB `m^(-2)` or tesla. | |
| 41931. |
If one of the two electrons of a H_(2) molecule is removed, we get a hydrogen molecular ion H_(2)^(+).In the ground state of an H_(2)^(+) , the two protons are separated by roughly 1.5 Å, and the electron is roughly 1 Å from each proton. Determine the potential energy of the system. Specify your choice of the zero of potential energy |
Answer» Solution :System of charges is shown in the figure  Charge on electron `q_(1)=- 1.6xx10^(-19)` C Charge on PROTON`q_(2)=q_(3)= 1.6xx10^(-19)` C Talcing potential zero at infinity distance then, potential energy of the system, `U = U_(12)+U_(23) +U_(13)` `= k[(q_(1)q_(2))/(r_(1))+(q_(2)q_(3))/(2)+(q_(1)q_(3))/(r_(3))]` `=9XX10^(9) [((-16xx10^(-19))(1.6xx10^(-19)))/(10^(-10))+((1.6xx10^(-19))^(2))/(1.5xx10^(-10))+(-(1.6xx10^(-19))^(2))/(10^(-10))]` `=(9xx10^(9)xx(1.6xx10^(19)xx1.6xx10^(-19))^(2))/(10^(-10))[-1+(1)/(1.5)-1]` `=(9xx10^(9)xx(1.6xx10^(19)xx1.6xx10^(-19))^(2))/(10^(-10))[(-3+1)/(1.5)]` `=23.04xx10^(-10)xx((-2)/(1.5))`J `=(-23.04xx10^(-19)xx2)/(1.6xx10^(-19)xx1.5)eV[because1.6xx10^(-19)` J = 1 eV] =-19.2 eV |
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| 41932. |
A trnsvese sinusoidal wave of amplitude a, wavelength lambda , and. frequency f is travelling on a stretched string. The maximum speed of any point on the string is v/10, where v is the speed of propagation of the wave. If a = 10^(-3) m and v = 10 m/s, then lambda and f are given by |
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Answer» `lambda = 2PI XX 10^(-2) m` |
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| 41933. |
Prove that, if an insulated, uncharged conductor is placed near a charged conductor and no other conductors are present, the uncharged body must be intermediate in potential between that of the charged body and that of infinity. |
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Answer» Solution :The electric field ` E =-(DV)/(dr)` suggests that electric potential decreases ALONG the DIRECTION of the field. LET us TAKE any path from the charged conductor to the uncharged conductor along the direction of electric field. Therefore, the electric potential decrease along this path. Now, another path from the uncharged conductor to infinity will again continually lower the potential further. This ensures that the uncharged body must be intermediate in potential between that of the charged body and that of infinity. |
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| 41934. |
Write the application of ICT ? |
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Answer» Solution :(i)Agriculture The implementation of information and communication technology (ICT) in agriculture sector enhances the productivity, improves the living standards of farmers and overcomes the challenges and risk factors. (a) ICT is widely used in increasing FOOD productivity and farm management (b) It helps to optimize the use of water, seeds and fertilizers ETC. (c) Sophisticated technologies that INCLUDE robots, temperature and moistureaerial IMAGES, and GPS technology can be used. (d) Geographic information systems are extensively used in farming to de suitable place for the species to be planted. (iii) Mining (a) ICT in mining improves operational efficiency, remote monitoring and dis locating system (b) Information and communication technology provides audio-visual warming trapped underground MINERS. (c) It helps to connect remote sites. |
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| 41935. |
(A) : When radius of circular loop carrying current is doubled, its magnetic moment becomes four times (R) : Magnetic rnoment depends on area of the loop |
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Answer» Both 'A' and 'R' are TRUE and 'R' is the CORRECT explanation of 'A'. |
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| 41936. |
Rydberg constant is: |
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Answer» a UNIVERSAL constant |
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| 41937. |
One end of each of two identical springs, each of force-constant 0.5N are attached on the opposite sides of a wooden block of mass 0.01Kg. The other ends of the springs are connected to separate rigid supports such that the springs are unstretched an are collinear in a horizontal plane. To the wooden piece is fixed a pointer which touches a vertically moving plane paper. The wooden piece, kept on a smooth horizontal table is now displaced by 0.02 m along the line of springs and released. If the speed of paper is 0.1 m/s. The distance between two consecutive maxima on this path is 6.28 xx.10^(-y)mtrs. Find y |
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Answer» |
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| 41938. |
Which of the following statements is not true?. |
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Answer» Solar cells MAY be future SOURCE of power for cars |
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| 41939. |
The time taken by sunlight to pass through a glass window (mu = 1.5) of thickness 4 mm is : |
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Answer» `2 XX 10^-11 s` |
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| 41940. |
An AM wave is represented by C_m (t)=6(1+0.5sin12560t)xxsin22xx10^5t. calculatefrequency of carrier wave. |
| Answer» SOLUTION :`omega_m=12560Hz` | |
| 41941. |
Unpolarised light is incident on a plane glass surface. What would be the angle of incidence so that the reflected and refracted rays are perpendicular to each other. |
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Answer» Solution :Here, reflected and refracted light rays are given perpendicular and so angle of incidence (1) must be equal to polarising angle or Brewster angle shown by symbol `theta_(p)`. Now, according to Brewsters. law, `tantheta_(P)=mu` `:.tantheta_(P)=1.5` `("":."Here "mu-mu_(ga)-(mu_(g))/(mu_(a))=(1.5)/(1)=1.5)` But `tan(56^(@)19.)=1.5003~~1.5` `impliestheta_(P)=56^(@)19.` NOTE: Here if `theta_(P)` is REQUIRED only in degree then `theta_(P)=56^(@)+19.=56^(@)+((19)/(60))^(@)` `=56^(@)+(0.32)^(@)` `=(56.32)^(@)` |
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| 41942. |
Two parallel plate capacitors X and Y have the same area of plates and same seperation and air between the plates while Y contains a dielectric medium of in_r=4 Calculate the potential difference between the plates of X and Y. |
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Answer» Solution :TOTAL CHARGE `Q = CV = 4muF xx15V = 60 MUC` `V_x =Q/C_x = (60 muC)/(5muF) = 12 V ` `V_x = Q / C_x =(60muC)/(5muF)=3V` |
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| 41943. |
A metal wire of mass 10 gm and length 10cm is placed on a smooth inclined plane of inclination 60° such that the length of wire is perpendicular of length of incline. If a current of 3A is passed through the wire vertically downward magnetic field applied for which the wire remains stationary on the inclined plane is (g = 10m//s^2) |
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Answer» `3T` |
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| 41944. |
Two parallel plate capacitors X and Y have the same area of plates and same seperation and air between the plates while Y contains a dielectric medium of in_r=4 Calculate capacitance of capacitor if equivalent of the combind system. |
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Answer» SOLUTION :`C_x=(4epsilon_0A)/(d)C = 4 C ` then `C_y=(4epsilon_0A)/(d)C = 4C ` For SERIES combination of Cx and CY `1/C=1/C_x+1/C_y` `implies1/4=1/C+1/(4C)=5/(4C)impliesC=5 MUF` `:.C_x=5muF and C_y=4xx5=20muF` |
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| 41945. |
Half-life of radioactive substance is 140 days. Initially, is 16g. Calculate the times for this substance when it reduces to 1 g : |
| Answer» Answer :D | |
| 41946. |
Burglar alarm is based on |
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Answer» THERMOELECTRICITY |
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| 41947. |
Two parallel plate capacitors X and Y have the same area of plates and same seperation and air between the plates while Y contains a dielectric medium of in_r=4 Estimate the ratio of electrostatic energy stored in X and Y. |
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Answer» SOLUTION :`U_x/V_y=((Q^2//2C_x))/((Q^2//2C_y))=C_x/C_y=(4C)/C =4/1` `impliesU_x : U_y=4:1` |
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| 41948. |
In a Cu-voltameter, mass deposited in 30 s is m gm. If the time-current graph is shown in the following figure What is the electrochemical equivalent of Cu? |
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Answer» m/2 `=1/2xx100xx10+100xx10+100xx10+1/2xx100xx10` `=3xx100xx10mAxxs` `=3000xx10^(-3)amps=3 C` W=ZQ `thereforeZ=W/Q=m/3` |
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| 41949. |
In Young's double-slit experiment, a point in the fringe pattern is one-third of the fringe width fr9m the central bright fringe. The phase difference between the interfering waves at the point is |
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Answer» `(pi)/(6)` rad `therefore" Phase difference "=(2pi)/(lamda)xxDelta =(2pi)/(3)` rad. |
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| 41950. |
What's the relation between B_axial and B_equtl ? |
| Answer» SOLUTION :`(B_(equtl))/(B_(AX IAL))` | |